Does every Lawvere theory arise in this way?

Question

By a Lawvere theory, I mean a finite-product category $\mathsf{T}$ equipped with a distinguished object, such that every object of $\mathsf{T}$ can be expressed as a finite product of the distinguished object.

Now let $X$ denote an object of a finite-product category $\mathbf{C}$. Then $X$ gives rise to a Lawvere theory $\mathrm{Lawv}(X)$ with distinguished object $X$, defined as the full subcategory of $\mathbf{C}$ induced by the objects $\{X^n \mid n:\mathbb{N}\}.$

Examples.

• Viewing $2$ as an object of $\mathbf{Set}$, we see that the arrows of $\mathbf{Lawv}(2)$ are "truth tables." Hence $\mathbf{Lawv}(2)$ is the Lawvere theory of Boolean algebras.

• Let $\mathbb{K}$ denote a field. Write $U_{\mathrm{Mod}}(\mathbb{K})$ for the underlying $\mathbb{K}$-module. Then $\mathbf{Lawv}(U_{\mathrm{Mod}}(\mathbb{K}))$ is the Lawvere theory of $\mathbb{K}$-modules.

Question. Is it true that for all Lawvere theories $\mathsf{T}$, there exists a Lawvere theory $\mathsf{S}$ and an $\mathsf{S}$-algebra $X$ such that $\mathrm{Lawv}(X) \cong \mathsf{T}$?

Answer 1

I can elaborate on the second example a bit more:

Let $T$ be a Lawvere theory. Then $T$ is commutative iff $T \cong Lawv(\text{hom}_T(x,-))$, where $x$ is the generic object in $T$.

A Lawvere theory is commutative if all its $n$-ary operations commute with each each other. A morphism of $T$-algebras is a natural transformation between functors; the naturality condition is precisely the requirement to commute with all $n$-ary operations. Your example of modules over a field (more generally a commutative semi-ring) is the prime example of a commutative theory.

Now commutative Lawvere theories are extremely useful, but also very restrictive. I will think about the general case some more.

Answer 2

The answer is yes for Lawvere theories which admit a "zero", that is, Lawvere theories in which there is a natural map $x\to y$ (natural in $x,y\in T$). This is equivalent to a map from the terminal $T$-algebra $*$ to the initial $T$-algebra; and happens for instance when the initial $T$-algebra is terminal which is the case for $R$-modules.

This is a consequence of the following general observation:

For any small finite-product category $C$ in which there exists a transformation $x\to y$, natural in $x,y\in C$, there is a product-preserving embedding from $C$ into $Mod_S$ for some $S$.

Applying this to $C= T$, we get the desired result, so let me explain the proof of that.

Fix any such $C$, and consider the object $ Fun^\times(C,Set)$ given by $F:= \prod_{c\in C} \hom(c,-)$. The full subcategory $S\subset Fun^\times(C,Set)$ spanned by $F^{\times n}, n \in \mathbb N$ is a Lawvere theory with distinguished object $F$, and I claim $C$ embeds in a product-preserving way into $Mod_S$ - the embedding is given by $C\to Fun^\times(S,Set), c\mapsto (G\mapsto G(c))$.

(in terms of algebras, I am claiming that for a given Lawvere theory $T$, $S$ can be chosen as $\mathbf{Lawv}(\prod_n Fn)$, where $F$ is the free $T$-algebra functor)

This is clearly a functor, so now it suffices to see that it is in fact fully faithful. A map $f: d\to d'$ induces $\prod_{c\in C}\hom(c,d)\to\prod_{c\in C}\hom(c,d')$. Since all the terms in the product are nonempty, it follows that one can recover the component $\hom(d,d)\to \hom(d,d')$ from this map and thus it is faithful.

Now, let $\alpha_G: G(d)\to G(d')$ be a morphism, natural in $G\in S$. I claim that it must come from some map $f :d\to d'$.

Fix some natural morphism $0_{x,y}: x\to y$. For any $d\in C$, consider the following morphism $F\to F$- $\prod_{c\in C}\hom(c,-)\to \prod_{c\in C}\hom(c,-)$ : on the $c$-coordinate of the target, pick out $0_{c,-}\in \hom(c,-)$ if $c\neq d$, and on the $d$-coordinate, just pick the projection $F\to \hom(d,-)$. This is a morphism of functors.

Naturality of $\alpha$ forces it to commute with this, and it follows that $\alpha_F$ is a product of maps $\hom(c,d)\to \hom(c,d')$. The claim is now that these are natural in $c$, and this can be proved in a similar way, by considering, for every $g:c_0\to c_1$ the map $F\to F$ which is the identity on most coordinates, and which on the coordinate $\hom(c_0,-)$ first projects onto $\hom(c_1,-)$ and then precomposes by $g$.

As it is natural in $c$, it forces it to be given by $f\circ -$ for some (unique) $f: d\to d'$.

Now for a general $G= F^{\times n}\in S$, you can use the projections $G\to F$ and naturality of $\alpha$ to deduce that the whole of $\alpha$ is in fact given by $G(f): G(d)\to G(d')$, which proves that our functor was, in fact, fully faithful.